# Compute distribution for count distribution

In Poisson point processes, we first let $$N$$ be a random Poisson random variable, with parameter $$\lambda$$: $$P(N=n)=\frac{e^{-\lambda}\cdot\lambda^n}{n!},\qquad n=0,1,\ldots$$ Without loss of generality, we let $$Y = (Y_1, Y_2, \dots, Y_n,\dots)$$ be an i.i.d. sequence (infinite) of standard normal random variables. In addition, assume $$N$$ is independent of $$Y$$. $$f_{Y_i}(y_i)=\frac{1}{\sqrt{2\pi}}e^{-\frac{y_i^2}{2}},\qquad -\infty Now we take an arbitrary interval $$[a, b]$$ and let $$X$$ be the number of $$Y_i$$s landing on the interval $$[a,b]$$, I want to know what is the distribution of $$X$$ here, I know it is a little bit complicated...

I searched Wikipedia and read the section on the uniformity of the points, but I still not getting it.

I think you're confusing things here because this has nothing to do with the distribution of points of the Poisson Process. Fix $N$ for a moment. Then the number of $Y_i$'s that fall within $[a,b]$ is binomially distributed: define $p:=p({[a,b]})=P(Y_1\in [a,b])=F(b)-F(a)$, where $F$ is the CDF of the standard normal. Then $X$ is binomially distributed (conditional on $N$ fixed): $P(X|N)=\binom{N}{X}p^X(1-p)^{N-X}$.
You're interested in $N$ being Poisson distributed. Since you're allowed $N=0$ with positive probability, lets define $P(X=0|N=0)=1$. Above we calculated the conditional distribution of $X$ so:
$$P(X)=\sum_{N=0}^\infty P(X|N)P(N),$$
• Thanks so much for your help. It is really a nice point that we define p as the probability that $Y_i$ is in [a,b] so $p = F(b)-F(a)$ But I still wonder whether the binomial result is the probability mass function, or the conditional distribution..... Again thanks so much! Dec 1 '17 at 18:59
• I'm not sure I understand your question. $X$ is binomially distributed, conditional on the value of $N$. Dec 1 '17 at 21:04